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Liono4ka [1.6K]
2 years ago
10

Kelly blends coffee. She mixes brand A costing $6 per kilogram with brand B costing $8 per kilogram. How many kilograms of each

brand does she have to mix to make 50 kg of coffee costing her $7.20 per kg?​
Mathematics
1 answer:
sukhopar [10]2 years ago
6 0

Step-by-step explanation:

Let

x

be the kg of coffee of brand A in the mix and

y

be the kg of coffee of brand B in the mix.

The total kg must be

50

.

x

+

y

=

50

The cost per kg of the mix must br

$

7.20

. For this, the total cost of the mix will be

6

x

+

8

y

, so the total cost per kg of the mix will be

6

x

+

8

y

50

.

6

x

+

8

y

50

=

7.20

Now that we have our two equations, we can solve.

6

x

+

8

y

=

7.20

⋅

50

6

x

+

8

y

=

360

From the first equation, we can multiply both sides by

6

to get:

6

x

+

6

y

=

300

Subtracting, we get:

2

y

=

60

y

=

30

Thus, we need

30

kg of brand B in our mix. This means that

50

−

30

=

20

kg will be of brand A.

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X2 + 2y ÷ w + 3z for w = 2, x = 5, y = 8, and z = 3?
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Step-by-step explanation:

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Identify the functions that are continuous on the set of real numbers and arrange them in ascending order of their limits as x t
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Answer:

g(x)<j(x)<k(x)<f(x)<m(x)<h(x)

Step-by-step explanation:

1.f(x)=\frac{x^2+x-20}{x^2+4}

The denominator of f is defined for all real values of x

Therefore, the function is continuous on the set of real numbers

\lim_{x\rightarrow 5}\frac{x^2+x-20}{x^2+4}=\frac{25+5-20}{25+4}=\frac{10}{29}=0.345

3.h(x)=\frac{3x-5}{x^2-5x+7}

x^2-5x+7=0

It cannot be factorize .

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2.g(x)=\frac{x-17}{x^2+75}

The denominator of g is defined for all real values of x.

Therefore, the function g is continuous on the set of real numbers

\lim_{x\rightarrow 5}\frac{x-17}{x^2+75}=\frac{5-17}{25+75}=\frac{-12}{100}=-0.12

4.i(x)=\frac{x^2-9}{x-9}

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x=9

The function i is not defined for x=9

Therefore, the function i is  not continuous on the set of real numbers.

5.j(x)=\frac{4x^2-7x-65}{x^2+10}

The denominator of j is defined for all real values of x.

Therefore, the function j is continuous on the set of real numbers.

\lim_{x\rightarrow 5}\frac{4x^2-7x-65}{x^2+10}=\frac{100-35-65}{25+10}=0

6.k(x)=\frac{x+1}{x^2+x+29}

x^2+x+29=0

It cannot be factorize .

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\lim_{x\rightarrow 5}\frac{x+1}{x^2+x+29}=\frac{5+1}{25+5+29}=\frac{6}{59}=0.102

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x^2-8x-x+8=0

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(x-8)(x-1)=0

x=8,1

The function is not defined for x=8 and x=1

Hence, function l is not  defined for all real values.

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The denominator of m is defined for all real values of x.

Therefore, the function m is continuous on the set of real numbers.

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